Suppose that $w$ is a covector field on $[a,b]$, then $w$ can be written as $f(x)dx$, where $dx$ is the differential of the identity function. Define $\int_{[a.b]}w = \int_{[a,b]}f(x)\ dx$. This question is a little stupid, but what does $\int_{[a,b]}f(x)\ dx$ mean here?
I mean if I view $f(x)\ dx$ here as a differential, then it would take value in $[a,b]$ and produce a value in the cotangent bundle of $R$. However, this is not an object that I can integrate in the usual sense. But if I view $\int_{[a,b]}f(x)\ dx$ as the usual Riemann integral(where dx is only a notation) rather than view the integrand as a differential, then the integral will make sense. Which approach should I take here?
I'm sorry for being brief in my comment, I am a bit tired but let me just expand on a few things. In first-year calculus we get a really good geometric interpretation of the integral namely, the $n$-volume bounded by the graph of a function and the plane $x_n = 0$. As it stands, integration on a arbitrary space $X$ would not make sense unless we had a way to "coordinatize" it's points. Well for manifolds we have such maps. In the case of manifolds, you think immediately, "if you wanted to know the integral about some point $p \in U \subset M$, how could we connect $U$ to a subset of $\mathbb{R}^n$? Well, use the charts $(\phi,U)$ in the atlas of your manifold. You have $\phi: U \to \phi(U):=V \subset \mathbb{R}^n$ and so $\phi_*: T_pU \to T_{\phi(p)}\phi(U)$ which implies $\phi^*: T_{\phi(p)}\phi(U) \to T_pU$. Hence, you would like to have;
$$(*) \ \ \int_U \omega = \int_{\phi(U)} (\phi^{-1})^* \omega$$
which turns out to be the right definition so long as $M$ is orientable. Recall that if $U$ is already a subset of Euclidean space, say $\mathbb{R}^n$ then we define:
$$\int_U f \ dx^1 \wedge \cdots \wedge dx^n = \int_U f \ dx^1 \ dx^2 \cdots dx^n$$
and with this, $(*)$ becomes the perfect generalization for integration on orientable manifolds $M$ which are not just subsets of Euclidean space.